aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
Answer: 6x -9
Step-by-step explanation:
Answer:
150.884615385
Step-by-step explanation:
The normal distribution curve for the problem is shown below
We need to standardise the value X=405.5 by using the formula
![z-score= \frac{X-Mean}{Standard Deviation}](https://tex.z-dn.net/?f=z-score%3D%20%5Cfrac%7BX-Mean%7D%7BStandard%20Deviation%7D%20)
![z-score= \frac{405.5-402.7}{8.8} =0.32](https://tex.z-dn.net/?f=z-score%3D%20%5Cfrac%7B405.5-402.7%7D%7B8.8%7D%20%3D0.32)
We now need to find the probability of z=0.32 by reading the z-table
Note that z-table would give the reading to the left of z-score, so if your aim is to work out the area to the right of a z-score, then you'd need to do:
![P(Z\ \textgreater \ z)=1 - P(Z\ \textless \ z)](https://tex.z-dn.net/?f=P%28Z%5C%20%5Ctextgreater%20%5C%20z%29%3D1%20-%20P%28Z%5C%20%5Ctextless%20%5C%20z%29)
from the z-table, the reading
![P(Z\ \textless \ 0.32)](https://tex.z-dn.net/?f=P%28Z%5C%20%5Ctextless%20%5C%200.32%29)
gives 0.6255
hence,
![P(Z\ \textgreater \ 0.32)=1-0.6255=0.3475](https://tex.z-dn.net/?f=P%28Z%5C%20%5Ctextgreater%20%5C%200.32%29%3D1-0.6255%3D0.3475)
The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%